1. ## equation of hyperbola

Hi! I need help to solve a task.

Write an equation of hyperbola. Foci coincides with the ellipse (x^2/35+y^2/10=1) focal points. There is given also a point from hyperbola - L(5sqrt(2);4)

2. ## Re: equation of hyperbola

Originally Posted by Kiiefers
Hi! I need help to solve a task.

Write an equation of hyperbola. Foci coincides with the ellipse (x^2/35+y^2/10=1) focal points. There is given also a point from hyperbola - L(5sqrt(2);4)

1. With an ellipse the foci have the coordinates $F_1(-e,0)~,~F_2(e,0)$ with $e = \sqrt{a^2-b^2}$
That means the foci are at $F_1(-5,0)$ and $F_2(5,0)$

2. A hyperbola is defined by a constant difference of the distances between a point L on the hyperbola and the foci:

$|LF_1|-|LF_2| = 2a$

with $L(5\sqrt{2},4)$

I've got $a = \frac12 \cdot \sqrt{182-2\sqrt{3281}}$

3. The excentricity of a hyperbola is calculated by:

$e = \sqrt{a^2+b^2}$

Since the excentricity is e = 5 you can determine the value of b.

I've got $b=\frac12 \cdot \sqrt{2\sqrt{3281}-82}$

(For this part I used my computer)

4. Therefore the equation of the hyperbola is:

$\frac{x^2}{\left( \frac12 \cdot \sqrt{182-2\sqrt{3281}} \right)^2}-\frac{y^2}{\left( \frac12 \cdot \sqrt{2\sqrt{3281}-82} \right)^2}=1$

5. I've attached a drawing of the ellipse and the hyperbola with the point L on the right branch.

3. ## Re: equation of hyperbola

Thanks for help.